Question

2(y+z)dx-(x+z)dy+(2y-x+z)dz=0

Answer 1

Solution:

Given: $2(y+z)dz-(x+z)dy+(2y-x+z)dz=0$

We know that the general form of the Pfaffian equation is

$P d x+Q d y+R d z=0$

The integrability condition for the Pfaffian equation is

$(\operatorname{curl\ F}, F)=0$

where $F=(P, Q, R)$ , or

$P\left(\frac{\partial Q}{\partial z}-\frac{\partial R}{\partial y}\right)+Q\left(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}\right)+R\left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}\right)=0$

Here, $P=2(y+z), Q=-(x+z), R=(2 y-x+z)$

Now, verification: LHS

$=2(y+z)(-1-2)-(x+z)(-1-2)+(2 y-x+z)(2+1)$

$=-6 y-6 z+3 x+3 z+6 y-3 x+3 z\\=0$

$=$ RHS

Hence, verified.

So, the integrability condition for this equation hold.

If the Pfaffian equation is multiplied by a certain function $\mu(x, y, z)$ then one can obtain in the

left-hand side the total differential.

Next, if treating $z$ as a constant, then the given equation reduces to

$2(y+z) d x-(x+z) d y=0 \\$

$\Rightarrow\frac{2}{x+z} d x-\frac{1}{y+z} d y=0$

which has a solution

$U(x, y, z)=\frac{(x+z)^{2}}{y+z}=C_{1}$

The integrating factor $\mu(x, y, z)$ is given by

$\mu=\frac{1}{P} \cdot \frac{\partial U}{\partial x}=\frac{1}{2(y+z)} \cdot \frac{2(x+z)}{y+z}=\frac{x+z}{(y+z)^{2}}$

Then, we get

$\begin{array}{c} K=\mu R-\frac{\partial U}{\partial z}= \\ =\frac{x+z}{(y+z)^{2}}(2 y-x+z)-\frac{2(x+z)(y+z)-(x+z)^{2}}{(y+z)^{2}}= \\ =\frac{x+z}{(y+z)^{2}}(2 y-x+z-2 y-2 z+x+z)=0 \end{array}$

Thus we have the equation

$d U=0$

which has a solution $U=c$

Thus the solution of the given equation is

$(x+z)^{2}=c(y+z)$

where $c$ is a constant.